Gravitational Velocity of Mass ##M## at Distance ##d## from Gravitating Body

[ChrisVer]

Starting from a locked thread I tried to work the gravity of a body of mass $M$ on another body starting from infinity to some distance $d$ from the gravitating body.
We have from the SR 2nd Newton law that:
\gamma^3 a = \frac{GM}{r^2}

writting a= \frac{dv}{dt}= v \frac{dv}{dr} = \frac{1}{2} \frac{dv^2}{dr}

Naming v^2/c^2= x the above relation becomes:

\frac{dx}{(1-x)^{3/2}}= \frac{2GM}{c^2 r^2}dr

Integrating:

\int_{0}^{x_d}\frac{dx}{(1-x)^{3/2}}= \int_{\infty}^{d} \frac{2GM}{c^2 r^2}dr
\frac{2}{\sqrt{1-x_d}}-2= - \frac{2GM}{c^2 d}

x_d=1- \Big( 1- \frac{GM}{c^2 d}\Big)^{-2}

And so:

v(d)= c \bigg[ 1- \Big( 1- \frac{GM}{c^2 d}\Big)^{-2} \bigg]^{1/2}

I tried plotting this solution as v(d), the good part is that v<c for all distances however I don't understand why for d=GM/c^2 I'm obtaining an infinity (and worse- in the imaginary regime)? By the way, that's the Schwarzschild radius...

Even worse, if I set \frac{GM}{c^2}=1 the plot of \beta = v/c = \bigg[ 1- \Big( 1- \frac{1}{d}\Big)^{-2} \bigg]^{1/2} is given in my attachment...and doesn't seem to have a real solution away from 1?

 

[A.T.]

I tried plotting this solution as v(d), the good part is that v<c for all distances however I don't understand why for d=GM/c2 I'm obtaining an infinity

Are you trying to compute the falling speed relative to a local hovering observer? You cannot have one hovering at the Schwarzschild radius.

 

[ChrisVer]

. You cannot have one hovering at the Schwarzschild radius.

So the information of Schwarzschild radius singularity is contained within SR? I solved the problem in a mechanical way...
But the main problem is that for larger radii from the source (distances $d>R_s$), I am not getting a real solution for $v$...

 

[Mentz114]

Starting from a locked thread I tried to work the gravity of a body of mass $M$ on another body starting from infinity to some distance $d$ from the gravitating body.
We have from the SR 2nd Newton law that:
\gamma^3 a = \frac{GM}{r^2}

writting a= \frac{dv}{dt}= v \frac{dv}{dr} = \frac{1}{2} \frac{dv^2}{dr}

Naming v^2/c^2= x the above relation becomes:
..
..

Typo here maybe

$\frac{1}{2} \frac{dv^2}{dr}$

 

[ChrisVer]

Eh?
\frac{1}{2} \frac{dv^2}{dr}= \frac{1}{2} \big( \frac{dv}{dr} v + v \frac{dv}{dr}\big) = v \frac{dv}{dr}= \frac{dr}{dt} \frac{dv}{dr} = \frac{dv}{dt}=a

 

[A.T.]

We have from the SR 2nd Newton law that:

This old thread might be relevant:
https://www.physicsforums.com/threads/gravitational-acceleration-in-gr.310397/#post-2177809

 

[ChrisVer]

That looks like a GR conversation..

 

[A.T.]

That looks like a GR conversation..

Yes, for comparison of g under GR with the value you assume based on SR and Newton.

 

[ChrisVer]

I think this is a mistake with the gravitational force I used...I think I should have used a minus sign: \gamma^3 a = - GM/r^2...
The main problem is that, as it is, there is no solution for d &gt; R_s (v^2&lt;0) and this makes physically no sense, since the force I used could as well be some other 1/r^2 type force (like electromagnetism), with the change R_s \rightarrow D=\frac{k q_1 q_2}{c^2 m}.

 

[ChrisVer]

The result for v is then:

v= c \sqrt{1- (1+\frac{GM}{c^2 d} )^{-2}}
With a plot (for R_s= \frac{GM}{c^2}=1) like the attached, which makes "sense".

 

[Mentz114]

Eh?
\frac{1}{2} \frac{dv^2}{dr}= \frac{1}{2} \big( \frac{dv}{dr} v + v \frac{dv}{dr}\big) = v \frac{dv}{dr}= \frac{dr}{dt} \frac{dv}{dr} = \frac{dv}{dt}=a

Oh, you mean $\frac{d}{dr}v^2$. Your notation confused me.

What is the plot of in your latest post ?

 

[ChrisVer]

What is the plot of in your latest post ?

It is let's say \frac{v}{c} = \beta(r)... it shows the change of the velocity of a mass that began from infinity with 0 velocity under the "Newtonian" gravitational law of a point-like mass source M (in terms of "schwarzchild radii" of the source body- of course it should not be confused with schwarzchild radii since I don't work in GR). So for example, when the body is at a distance x=2 \times \frac{GM}{c^2} away from the source, it will have a velocity v \approx 0.75c...
In a closed thread someone said that the velocity of the object would eventually reach a value larger than c due to acceleration. I just wanted to counter this idea via special relativity alone...

I guess this could change for GR, but OK... It was more like a special relativistic application of F= \frac{dp}{dt}. Showing that by applying special relativity alone, it doesn't matter for how long your body has traveled (starting from infinity up to reaching the source at 0), its velocity won't exceed the speed of light.

 

[ChrisVer]

It saves the "classical"/non-relativistic prediction that:
v= \sqrt{\frac{2GM}{r}}
Which allows for r such that v&gt;c, eg r= \frac{2GM}{9c^2} which gives v=3c

 

[Mentz114]

It is let's say \frac{v}{c} = \beta(r)... it shows the change of the velocity of a mass that began from infinity with 0 velocity under the "Newtonian" gravitational law of a point-like mass source M [...] So for example, when the body is at a distance x=2 \times \frac{GM}{c^2} away from the source, it will have a velocity v \approx 0.75c...
In a closed thread someone said that the velocity of the object would eventually reach a value larger than c due to acceleration. I just wanted to counter this idea via special relativity alone...

The curve certainly makes sense. You can see those $\gamma$'s kicking in.

 

[PeterDonis]

We have from the SR 2nd Newton law

SR can't be used to model gravity.

So the information of Schwarzschild radius singularity is contained within SR?

No. You need to use GR to model gravity.

That looks like a GR conversation..

So should this thread be. You can't model gravity using SR.

 

[PeterDonis]

It was more like a special relativistic application of $F= \frac{dp}{dt}$.

But for gravity, $F = 0$. Gravity is not a force in relativity. A force in relativity is something that causes proper acceleration. Gravity doesn't. However the computations you are making work out formally, they are physically meaningless as far as gravity is concerned.

 

[Dale]

Starting from a locked thread

Please do not reopen locked threads.